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3 years ago

Which type of vehicle is used routinely at construction and mining sites?

Engineering

2 answers:

Semmy [17]3 years ago

6 0

Answer:

Hydraulics as the name implies is pertaining to water. Hydraulic vehicle is a vehicle used on the construction site. The vehicle makes use of liquids to function. The fluid powers the engine of the vehicle do as to generate power.

Types of fluids used in hydraulic vehicles includes: Water fluid, petroleum fluid, synthetic fluid.

Explanation:

it really depends on the job.

satela [25.4K]3 years ago

5 0

We are required to state the type of vehicle which is routinely used at the construction site.

The vehicle which is routinely used at the construction site is called hydraulic vehicle.

Hydraulics as the name implies is pertaining to water. Hydraulic vehicle is a vehicle used on the construction site. The vehicle makes use of liquids to function. The fluid powers the engine of the vehicle do as to generate power.

Types of fluids used in hydraulic vehicles includes: Water fluid, petroleum fluid, synthetic fluid.

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3 years ago

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Acertain foundation will experience a bearing capacity failurewhen it is subjected to a downward load of 2200 kN. Using ASD with

ehidna [41]

Answer:

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Explanation:

This is what I like to see teachers giving out.

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3 years ago

What’s the number of gold atoms in a nanogram? a picogram?

zvonat [6]

Answer :

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

The number of gold atoms in picogram is, $3.057\times 10^{9}$

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-9}$ nanograms of gold contains $\frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12}$ number of gold atoms

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-12}$ picograms of gold contains $\frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9}$ number of gold atoms

The number of gold atoms in picogram is, $3.057\times 10^{9}$

8 0

3 years ago

A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

$\sum M_A = 0$

-(P × 100)+ (W×35) = 0

P = $\frac{W*35}{100}$

Replacing 650 N for W

$P = \frac{650*35}{100}$

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

5 0

3 years ago

A sandy soil has a total unit weight of 120 pcf, a specific gravity of solids of 2.64, and a water content of 16 percent. Comput

olchik [2.2K]

Answer:

A). Dry unit weight = 1657.08Kg/m3

B). Porosity = 0.37

C). Void ratio = 0.593

D). 0.712

Explanation:

Total unit weight, Y = 120pcf =1922.2 Kg/m3

Specific gravity of solids, Gs = 2.64

Water content, w = 16%

A). Dry unit weight

Yd = Y/(1+w)

= 1922.2/(1+0.16) = 1657.08Kg/m3

B). Porosity

However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3

Void ratio = 2.64×1000/1657.08 = 0.593

And porosity = e/(1+e) =0.593/(1+0.593) = 0.37

C). void ratio, e = 0.593

D). Degree of saturation, S = m×Gs/e where m =water content

S = 0.16×2.64/0.593 = 0.712

5 0

3 years ago